How many number plates can be made if the number plates have two letters of the English alphabet $(a - z)$ followed by two digits $(0 - 9)$ if the repetition of digits or alphabets is not allowed?

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Solution

Solve for the number of combinations of number plates
No. of English alphabets $(a - z) =$ $26$
No. of digits $(0 - 9) =$ $10$
We have total four place to cover with two alphabets and two digits.
_ _ _ _
(alphabet) (alphabet) (digit) (digit)
For the first place:
Out of $26$ alphabets we need one. Therefore, combinations for first place is
${}^{26}C_{1} = \frac{26!}{1! \times (26 - 1)!} = \frac{26!}{25!} = \frac{26 \times 25!}{25!} = 26$
For the second place:
We are left with $25$ alphabets now as no repetitions are allowed.
Therefore combinations for second place is
${}^{25}C_{1} = \frac{25!}{1! \times (25 - 1)!} = \frac{25!}{24!} = \frac{25 \times 24!}{24!} = 25$
For the third place:
Out of $10$ digits we need to select one . Therefore combinations for third place is
${}^{10}C_{1} = \frac{10!}{1! \times (10 - 1)!} = \frac{10!}{9!} = \frac{10 \times 9!}{9!} = 10$
For the fourth place:
We are left with $9$ digits now as no repetitions are allowed. Therefore combinations for fourth place is
$^{9} C_{1} = \frac{9!}{1! \times (9 - 1)!} = \frac{9!}{8!} = \frac{9 \times 8!}{8!} = 9$
In order to find the total no. of possible number plate that can be made, we will multiply the combinations for each of the places. So total Combinations of number plates $= 26 \times 25 \times 10 \times 9 = 58500$
Hence, the required number of plates is $58500$

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How many number plates can be made if the number plates have two letters of the English alphabet a-z followed by two digits (0-9)if the repetition of digits or alphabets is not allowed?

How many number plates can be made if the number plates have two letters of the English alphabet $(a - z)$ followed by two digits $(0 - 9)$ if the repetition of digits or alphabets is not allowed?